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\begin{document}
\title{Math 525: Lecture 18}
\date{March 15, 2018}
\maketitle
Consider a gambler playing roulette. He has decided that he will place
a bet on 17 at time $\tau$. We can model the game of roulette as
a stochastic process with $X_{n}$ being the result of the $n$-th
spin. In a perfect world, the gambler would choose
\[
\tau=\inf\{n\geq0\colon X_{n+1}=17\},
\]
the time just before the ball lands on 17. Of course, this is cheating!
It would mean that the gambler can look into the future. Indeed, the
gambler can base his decision to bet at time $\tau=n$ only on \emph{information
available at time $n$} (e.g., the results of the spins at the previous
times $X_{0},X_{1},\ldots,X_{n}$). But how can we express the statement
``information available before time $n$'' mathematically?
\section{Filtrations}
For our discussion, let $(X_{n})_{n\in T}$ be a Markov chain on a
countable state space $S\subset\mathbb{R}$. We will not assume that
the state space $S$ is finite, nor that the Markov chain is stationary.
Note that if $T$ is finite, $S^{T}$ is countable and hence we can
take $(\Omega=S^{T},\mathcal{F}=2^{\Omega},\mathbb{P})$ to be the
probability space (for some $\mathbb{P}$).
\begin{defn}
Let $\mathcal{F}$ be a $\sigma$-algebra and $T$ be an index set.
A \emph{filtration} is a sequence $(\mathcal{F}_{t})_{t\in T}$ of
$\sigma$-algebras such that $\mathcal{F}_{s}\subset\mathcal{F}_{t}\subset\mathcal{F}$
for all $s\leq t$.
\end{defn}
While the above definition is general enough for continuous index
sets, we will focus on the case when $T$ is discrete (e.g., $T=\{0,1,\ldots\}$
or $T=\{0,1,\ldots,N\}$) since we are motivated by Markov chains.
The following illustrates that $\mathcal{F}_{t}$ can be thought of
as all information accrued up to time $t$.
\begin{example}
A coin is flipped twice. This corresponds to the sample space
\[
\Omega=\{HH,HT,TH,TT\}.
\]
Since this a finite sample space, we can take $\mathcal{F}=2^{\Omega}$
to be our $\sigma$-algebra. Define
\begin{align*}
\mathcal{F}_{0} & =\{\emptyset,\Omega\}\\
\mathcal{F}_{1} & =\{\emptyset,\Omega,\{HH,HT\},\{TH,TT\}\}\\
\mathcal{F}_{2} & =2^{\Omega}.
\end{align*}
Note that $(\mathcal{F}_{n})_{n\in\{0,1,2\}}$ defines a filtration
since $\mathcal{F}_{0}\subset\mathcal{F}_{1}\subset\mathcal{F}_{2}=\mathcal{F}$.
Now, let $\omega=(\omega_{1},\omega_{2})\in\Omega$ be the result
of the coin flips. After we have observed the first coin flip, we
are able to answer whether
\[
\omega\in\{HH,HT\}
\]
(i.e., whether the first coin flip was heads). However, we cannot
answer whether
\[
\omega\in\{HH\}
\]
since this would require intimate knowledge of the second coin flip.
More generally, at time $n$, we can answer any question of the form
\[
\omega\in F\qquad\text{where }F\in\mathcal{F}_{n}.
\]
In this sense, $\mathcal{F}_{0}$ is the ``dumb'' sub$\sigma$-algebra,
since it carries no information whatsoever!
\end{example}
As you might have noticed in the above example, it is quite cumbersome
to write out filtrations by hand (and our example only had two coin
flips!) There is a better way:
\begin{defn}
We say a random variable $X$ is \emph{measurable} with respect to
some $\sigma$-algebra $\mathcal{G}$ (or simply ``$\mathcal{G}$
measurable'') if
\[
X^{-1}(B)\in\mathcal{G}\qquad\text{for }B\in\mathcal{B}(\mathbb{R}).
\]
Let $\{X_{\alpha}\}_{\alpha}$ be family of random variables on the
same probability space. Then \emph{the $\sigma$-algebra generated
by the family} is the smallest $\sigma$-algebra $\mathcal{G}$ such
that for each $\alpha$, $X_{\alpha}$ is $\mathcal{G}$ measurable.
We will often denote this $\sigma$-algebra by $\sigma(\{X_{\alpha}\})\equiv\mathcal{G}$.
\end{defn}
If the family $\{X_{0},\ldots,X_{n}\}$ is finite, we'll forgo the
curly braces and write $\sigma(X_{0},\ldots,X_{n})$.
\begin{example}
Returning to the coin flipping example, let
\[
X_{n}=\begin{cases}
1 & \text{if the }n\text{-th coin flip is heads}\\
0 & \text{otherwise}.
\end{cases}
\]
Let $B$ be a Borel set. Note that there are four possibilities:
\begin{itemize}
\item $1\in B$ and $0\notin B$ in which case $X_{1}^{-1}(B)=\{HH,HT\}$,
\item $0\in B$ and $1\notin B$ in which case $X_{1}^{-1}(B)=\{TH,TT\}$,
\item $0,1\in B$ in which case $X_{1}^{-1}(B)=\Omega$.
\item $0,1\notin B$ in which case $X_{1}^{-1}(B)=\emptyset$, and
\end{itemize}
Therefore,
\[
\sigma(X_{1})=\mathcal{F}_{1}=\{\emptyset,\Omega,\{HH,HT\},\{TH,TT\}\}.
\]
Similarly, you can check that $\sigma(X_{1},X_{2})=\mathcal{F}_{2}$.
\end{example}
\begin{defn}
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and
$\mathbb{F}=(\mathcal{F}_{t})_{t\in T}$ be a filtration on $\mathcal{F}$
where $T$ is some index set. Then, $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$
is called a \emph{filtered probability space}.
\end{defn}
\section{Stopping times}
For the remainder of this section, we will assume a filtered probability
space $(\Omega,\mathcal{F},\mathbb{F}=(\mathcal{F}_{t})_{t\in T},\mathbb{P})$
with index set $T=\{0,1,\ldots\}$ or $T=\{0,1,\ldots,N\}$ (though
the ideas extend easily to general index sets). We will also assume
a Markov chain on this space (i.e., $(X_{n})_{n\in T}$).
Armed with the notion of a filtration, we can try to outline some
fair rules that prevent our greedy gambler from being able to ``look
into the future''.
\begin{defn}
$\tau$ is a \emph{stopping time} if
\begin{enumerate}
\item it is a random variable taking values in $T\cup\{\infty\}$ and
\item for each $n\in T$, $\{\tau=n\}\equiv\{\omega\colon\tau(\omega)=n\}$
is a set in $\mathcal{F}_{n}$.
\end{enumerate}
\end{defn}
The first property just tells us that $\tau$ is a random time (in
the roulette example, $\infty$ corresponds to the gambler choosing
to never place his bet!). The second property tells us that the decision
to ``stop'' at time $n$ (e.g., place the bet at time $n$) can
only depend on the information witnessed up to time $n$.
\begin{prop}
A random variable $\tau$ taking values in $T\cup\{\infty\}$ is a
stopping time if and only if
\begin{equation}
\{\tau\leq n\}\in\mathcal{F}_{n}\qquad\text{for each }n\in T.\label{eq:alternate_stopping_time}
\end{equation}
\end{prop}
\begin{proof}
Suppose $\tau$ is a stopping time. Note that
\[
\left\{ \tau\leq n\right\} =\bigcup_{m=0}^{n}\{\tau=m\}.
\]
By definition, $\{\tau=m\}\in\mathcal{F}_{m}$ and $\mathcal{F}_{m}\subset\mathcal{F}_{n}$
whenever $m\leq n$. Therefore, $\{\tau\leq n\}$ is a finite union
of sets in $\mathcal{F}_{n}$, and hence $\{\tau\leq n\}\in\mathcal{F}_{n}$.
Suppose (\ref{eq:alternate_stopping_time}). Note that
\[
\left\{ \tau=n\right\} =\left\{ \tau\leq n\right\} \setminus\left\{ \taun\}$ are in $\mathcal{F}_{n}$.
\end{xca}
Let's see some examples of stopping times:
\begin{example}
~
\begin{enumerate}
\item The constant random variable $\tau=n$ is a stopping time since
\[
\left\{ \tau=m\right\} =\begin{cases}
\Omega & \text{if }m=n\\
\emptyset & \text{if }m\neq n.
\end{cases}
\]
\item Let $\mathcal{F}_{n}=\sigma(X_{0},\ldots,X_{n})$ for $n=0,1,2,\ldots$
where $(X_{n})_{n}$ is a sequence of random variables on the same
probability space. Let $B=(-1,1)$. Then,
\[
\tau_{B}=\inf\left\{ n\geq0\colon X_{n}\notin B\right\}
\]
where we use the convention $\inf\emptyset=\infty$. Then, $\tau_{B}$
is a stopping time, called the \emph{first exit time}. $B$ can be,
more generally, any Borel set. $\tau_{B}$ is a stopping time since
\begin{align*}
\left\{ \tau_{B}=n\right\} & =\left\{ X_{0}\in B,\ldots,X_{n-1}\in B,X_{n}\notin B\right\} .\\
& =\left\{ X_{0}\in B\right\} \cap\cdots\cap\left\{ X_{n-1}\in B\right\} \cap\left\{ X_{n}\notin B\right\} \\
& =X_{0}^{-1}(B)\cap\cdots\cap X_{n-1}^{-1}(B)\cap X_{n}^{-1}(B^{c})\\
& =\underbrace{X_{0}^{-1}(B)}_{\in\mathcal{F}_{0}}\cap\cdots\cap\underbrace{X_{n-1}^{-1}(B)}_{\in\mathcal{F}_{n-1}}\cap\underbrace{\left(X_{n}^{-1}(B)\right)^{c}}_{\in\mathcal{F}_{n}}.
\end{align*}
\item If $\tau$ is a stopping time and $m\geq0$ is an integer, $\tau+m$
is a stopping time. To see that this is a stopping time, note that
\[
\left\{ \tau+m=n\right\} =\left\{ \tau=n-m\right\} .
\]
\item If $\tau_{1}$ and $\tau_{2}$ are stopping times, so are $\tau_{1}\wedge\tau_{2}\equiv\min(\tau_{1},\tau_{2})$
and $\tau_{1}\vee\tau_{2}\equiv\max(\tau_{1},\tau_{2})$ (left as
an exercise).
\end{enumerate}
\end{example}
Given a stopping time $\tau$, we would like to define the value of
the Markov chain $(X_{n})_{n\in T}$ at time $\tau$. We do this below.
\begin{defn}
Let $\tau$ be a random variable taking values in $T$ (not necessarily
a stopping time). Define $X_{\tau}:\Omega\rightarrow\mathbb{R}$ by
\[
X_{\tau}(\omega)=X_{\tau(\omega)}(\omega).
\]
\end{defn}
\begin{prop}
$X_{\tau}$ defined above is indeed a random variable.
\end{prop}
\begin{proof}
This follows from the fact that
\[
X_{\tau}^{-1}(B)\equiv\left\{ \omega\colon X_{\tau}\in B\right\} =\bigcup_{n\geq0}\left(\left\{ \tau=n\right\} \cap\left\{ X_{n}\in B\right\} \right).
\]
\end{proof}
There is a potential problem: a stopping time $\tau$ is allowed to
take infinite values. However, we can still use the above to define
the value of the Markov chain at time $\tau$. For example,
\[
Y=X_{\tau}I_{\{\taun\right\} \right)^{c}\in\mathcal{F}_{n}.
\]
Lastly, let $\Lambda_{1},\Lambda_{2},\ldots\in\mathcal{F}_{\tau}$.
Then
\[
\bigcup_{n\geq0}\left(\Lambda_{n}\cap\left\{ \tau\leq n\right\} \right)=\left(\bigcup_{n\geq0}\Lambda_{n}\right)\cap\left\{ \tau\le n\right\} \in\mathcal{F}_{n}.
\]
\item This follows from
\[
\left\{ X_{\tau}I_{\{\tau